Interchange the deterministic and stochastic integrals

Question

We fix $T >0$ and let $\mathbb T$ be the interval $[0, T]$. Let $(X_t, t \in \mathbb T)$ be a continuous adapted process on some filtered probability space $(\Omega, \mathcal A, (\mathcal F_t)_{t \in \mathbb T}, \mathbb P)$. Let $g : \mathbb T \times \mathbb R^d \to \mathbb R$ be continuous. Assume that $$ \int_0^t g(s, X_s) \, \mathrm d s $$ is well-defined and $\mathbb P$-integrable for each $t \in \mathbb T$.

Can we interchange the deterministic and stochastic integrals, i.e., $$ \mathbb E \left [ \int_0^t g(s, X_s) \, \mathrm d s \right ] = \int_0^t \mathbb E \left [ g(s, X_s) \right ] \mathrm d s $$ ?

Thank you so much for your elaboration!

Answer 1

$\newcommand\om\omega\newcommand\Om\Omega$No. E.g., suppose that $\Om=(0,1]$, $T=1$, for each $t\in[0,T]$ the $\sigma$-algebra $\mathcal F_t=\mathcal A$ is the Borel $\sigma$-algebra over $\Om$, $\mathbb P$ is the uniform distribution over $\Om$, and $X_s(\om)=1/\om$ for $\om\in\Om$. Let $h$ be any continuous function from $[0,T]\times\mathbb R$ to $\mathbb R$ such that $$\int_1^\infty du\,\int_0^T ds\,h(s,u)\ne \int_0^T ds\,\int_1^\infty du\,h(s,u)$$ and $\int_1^\infty du\,\int_0^t ds\,h(s,u)\in\mathbb R$ for each $t\in[0,T]$. Let $g(s,u):=u^2 h(s,u)$ for all $(s,u)\in[0,T]\times\mathbb R$.

Then all your conditions hold, but $$\mathbb E\int_0^T ds\,g(s,X_s) =\int_1^\infty du\,\int_0^T ds\,h(s,u) \\ \ne\int_0^T ds\,\int_1^\infty du\,h(s,u) =\int_0^T ds\,\mathbb E g(s,X_s).$$